Question

Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components p_x , p_y and p_z . Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Answer 1

l_x = yp_z – zp_y
l_y = zp_x – xp_z
l_z = xp_y –yp_x
Linear momentum of the particle, p→ =\(\vec p =p_x\^i+p_y\^j+p_z\^k\)
Position vector of the particle, \(\vec r=x\^i+y\^j+z\^k\)
Angular momentum, \(\vec{l} =\vec{r} ×\vec{p} \)

\(= (x \^i + y \^j + z \^k) ×(p_x\^i+p_y\,\^J+p_z\^k)\)
\(\begin{bmatrix}  \^i&  \^j& \^k\\   x&  y& z\\   p_x&  p_y&  p_z\end{bmatrix}\)

\(l_x\^i+\^j+l_z\^k=\^i(yp_z-zp_y)-\^j(xp_z-zp_x)+\^k(xp_y-zp_x)\)

Comparing the coefficients of\(\^i,\^j\), and \(\^k\), we get :
\(\left.\begin{matrix}  l_x&  =yp_z& =zp_y\\   l_y&  =xp_z& -zp_x\\   l_z&  =xp_y&  yp_x\end{matrix}\right\}\) \(  …..(i)\)

The particle moves in the x-y plane. Hence, the z-component of the position vector and linear momentum vector becomes zero, i.e.,
z = p_z = 0
Thus, equation (i) reduces to :

 \(\left.\begin{matrix}  l_x=0&  & \\   l_Y=0&  & \\ l_z=xp_y-yp_x&  & \end{matrix}\right\}\)

Therefore, when the particle is confined to move in the x - y plane, the direction of angular momentum is along the z-direction.